Oxidation State Calculation
Oxidation State What is oxidation state? According to IUPAC "oxidation state gives the degree of oxidation of an atom in terms of counting electrons. The higher the oxidation state (OS) of a given atom, the greater is its degree of oxidation": '''OS of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds.''' When dealing with oxidation states we typically refer to them as '''''formal''''' oxidation states. As the definition says it is an '''''ionic approximation'''''. Real compounds are more complicated but the formal oxidation state is nonetheless a incredibly useful approximation. Why is oxidation state important? We are taught at an early age that oxidation state is important when understanding redox reactions. Whilst this is clearly true, it is also important across all chemical reactions. It allows us to simply understand the nature of chemical bonding by giving us a method to count electrons. It tells us how many atoms a particular atom is bonded to, and gives us information about the nature of the species involved. For example: FeCl2 & FeCl3 are different compounds with different properties and different reactivities. In the former the iron centre has an oxidation state of 2+ whilst in the latter it has an oxidation state of 3+. In inorganic chemistry we indicate the oxidation state with roman numerals. So iron (II) chloride and iron (III) chloride. Oxidation state has an enormous effect on reactivity. For example: In HCl the oxidation state of hydrogen is 1+. HCl is a strong acid as it readily releases H+ ions into solution. In NaH the oxidation state of hydrogen is 1-. NaH is a strong base as it readily releases H[[-|-]] ions into solution. Oxidation states are essential in inorganic chemistry, but are still incredibly important in organic chemistry. Calculating Oxidation States Rules are made to be broken Oxidation state is taught at a young age by using a variety of "rules". For example, in the UK these school level rules often include: * Hydrogen is always 1+ * Halogens are always 1- * Oxygen is always 2- * Alkali metals are always 1+ * Alkaline earth metals are always 2+ In this guide we are not interested in learning rules. We are going to use a method based entirely off the definition above and will let you figure out the oxidation state of any atom in any molecule just using the table below, a periodic table of electronegativities. For this we are using the Pauling scale (see below). You can do the same thing with the Allen, Mulliken, Allred-Rochow etc scales instead. Just make sure you use the same one for the same compound so that the relative values make sense. Always remember that as there are different values of electronegativity this system can give different results for elements that are similar in electronegativity, for example P-H bonds. ] O = Approach Approach Remember the definition of oxidation state? OS of an atom is the '''charge''' of this atom after '''''ionic approximation''''' of its '''''heteronuclear''''' bonds. Let's break it down. Oxidation state, by definition, only factors in '''''heteronuclear''''' bonds. Bonds to the same element contribute zero to the oxidation state. The elemental form of any given element is always zero regardless of the structure it takes. That includes all allotropes. O2? That's Oo. S8? That's So. Carbon in graphite, diamond, fullerenes? All Co. A lump of platinum? Pto. Neon gas? Neo. What is an ionic approximation of bonds? It means that we assume that all bonding is completely ionic, regardless of how covalent the bond is. So this is how we are going to calculate oxidation state. * Choose the atom you are interested in. * Break a bond between that atom and a different atom. * Assume that the electrons in each bond go to the most electronegative element * Balance the charges so we get an oxidation state contribution from that particular bond. * Repeat steps 2 to 4 until you have the contribution from all bonds. * Add the charges together for the atom you are interested in. This is much easier than it sounds. Let's go through a whole bunch of examples to help you fully understand this approach. Examples - Organic Hydrocarbons Methane Let's start with the simplest hydrocarbon, CH4. We are interested in the oxidation state of carbon so we have four C-H bonds to break, one at a time. C-H breaks to give us C and H. The electronegativity of carbon = 2.5 and hydrogen = 2.1. As carbon is the more electronegative element we give the electron pair to the carbon. Hydrogen has lost it's original electron so we have H+. We need to balances the charges so carbon has a C- charge. This is for one bond. As all four bonds are the same we end up with four H+ and four C-. Add all the charges together for carbon and we have the oxidation state, C4-. I hope you agree that the method was far easier than we expected. We can also apply it to the hydrogen atoms too. Each hydrogen has one bond to carbon. Break that bond, the electrons go to carbon leaving us with H+. Thus the oxidation state for hydrogen is H1+. Ethane Let's move on to the the next heaviest alkane, C2H6 (or H3CCH3)Sometimes it is more convenient to write chemical formulas in a way that shows the connectivity of the atoms you are interested in, like this. . We are still interested in the oxidation state of the carbon atoms so we have three C-H bonds and one C-C bond to break. As above C-H breaks to give us H+ and C-. There are three of them so we have C3-. Now we break the C-C bond. Remember that oxidation state is only affected by heteronuclear bonds! This homonuclear bond therefore contributes nothing to the oxidation state. In this bond we have C0. Add all the charges up and we have 3(C+)+(C0) which gives us a final oxidation state of C3-. Now apply this approach to ethene (H2C=CH2) and see what answer you get!You answer should be C2+ Oxygenated compounds Alcohols Let's move on to propanol, CH3CH2'''C'''H2OH. We are interested in the carbon attached to the oxygen. We have two C-H bonds, one C-C bond and one C-OH bond attached to that carbon. We have already established that we have C2- from the two C-H bonds (combined) and C0 from the C-C bond. That's C2- so far. Now we break the C-OH bond. We can see from our table that O (3.5) is significantly more electronegative than C (2.5). Unlike the C-H bond, it is now the oxygen that takes the electron pair. So we have C and -OH. Notice that it is important to factor in the bonds that the O has as well. We balance and now we have a C+ so C1+ from this bond. Add them all together, C2- + C0 + C1+ = C1- as the oxidation state for that carbon. Now let's look at propianaldehyde, CH3CH2CHO. We are still interested in the carbon attached to the oxygen. We have one C-H bond, one C-C bond and one C=O bond. As before we have C1- + C0 for the C-H and C-C bond. Now break the C=O bond. Treat it as two single bonds so oxygen (greedy git) gets both electron pairs. We have C and O2-. Balance and we have C2+ from this bond. Add them all together, C1- + C0 + C2+ = C1+ as the oxidation state for that carbon. '''This is why oxidation state is still so important in organic chemistry'''. The carbon at the business end of propanol has an ox. state of C1- whilst in the aldehyde it has an ox. state of C1+. As the carbon in the alcohol has to lose two electrons to become an aldehyde it is getting '''''oxidised'''''. If we want to do this transformation it means we have to have an oxidising agent (such as PCC). For the aldehyde to become an alcohol it needs to be reduced. This means we need a reducing agent (such as NaBH4). Throw NaBH4 at propanol in the hope of making the corresponding aldehyde and it will just look at you. Silently judging you. Examples - Inorganic Simple compounds NaBH4 Let's step it up a notch. Since we've just spoken about NaBH4 why not figure out the oxidation state of boron in that molecule. In this case we have four B-H bonds and a B-Na interaction. First the B-H bond. Boron (2.0) has an electronegativity slightly lower than that of hydrogen (2.1). So we break the B-H bond and, unlike in organic molecules, the electrons go to hydrogen so we have H-. Balance for boron and we get B+. That's a B4+ contribution from all the B-H bonds. What about the B-Na interaction? Here the approximation of the ionic from the get go and say an Na-B bond breaks to give Na and B. As B has a higher electronegativity than Na, B gets the electron pair to make Na+ and B-. B4+ + B- = B3+ for boron in NaBH4. H3PO4 Let's look at the oxidation state of phosphorus in phosphoric acid, H3PO4. If you've not seen this molecule you might want to make sure you can draw the Lewis structure of it before continuing. Done that? Good. In that case you will know that we have three P-OH bonds and one P=O bond. P (2.1) is less electronegative than O (3.5) so this example is just our alcohol and aldehyde example smooshed together. Breaking the P-OH bonds gives P+ and -OH. Breaking the P=O bond gives P2+ and O2-. Add them together, 3(P+) + P2+ = P5+ The reason I added this (relatively) simple compound is because of the bond breaking involved. The molecule is called phosphoric acid. It can therefore be confusing as to calculate the oxidation state you break the P-OH bonds to form -OH (the classic Bronsted base). Just remember this method has nothing to do with acid/base theory or anything else. KClO4 How about potassium chlorate, KClO3? You probably don't even need my help now, but let's figure out the oxidation state of the chlorine atom. We have three Cl=O bonds and a K-Cl bond to break. Oxygen (3.5) is more electronegative than chlorine (3.0) so it gets the electrons. Breaking the Cl=O bonds gives us O2- and Cl2+ for each bond. That's Cl6+ overall. Breaking the K-Cl bond gives us K+ and so Cl-. Add them to give 3(Cl2+) + Cl1- = Cl5+ A pretty good reason to not learn that halogens are always 1-! FO4F Yes you read that right, tetraoxygen difluoride, F-O-O-O-O-F. The oxidation state of the two central oxygens is clearly O0. For the end ones we have an O-O and a O-F bond. The O-O bond contributes nothing to oxidation state. Fluorine (4.0) is more electronegative than O (3.5) so breaking the F-O bond makes F-. To balance O must be O+. The oxidation state of oxygen is O1+, not the commonly quoted "O is always 2-" value! Inorganic Complexes I can almost hear you yawning. Let's step it up again. Iron pentacarbonyl What is the oxidation state of iron in Fe(CO)5? Well let's start by breaking the Fe-CO bond. Carbon is more electronegative so the electrons go to carbon giving us -CO. Iron is therefore Fe+ for each bond right? Wrong! CO isn't -CO but rather -CO+. Overall carbon monoxide is a neutral species. Neutral ligands bond in a dative fashion (i.e. donating a lone pair). As the metal centre doesn't have to give up an electron to this bond, it contributes nothing to the oxidation state. Therefore the formal oxidation state is Fe0. [Rh(COD)Cl]2 Moving on we now have the chloro(1,5-cyclooctadiene [i.e. COD])rhodium dimer. Here we have a different species for two reasons. The first is that one of the ligands is a bidentate alkene donor and the second that we are dealing with a dimer. To deal with a dimer we simply cut it in half and deal with the monomeric species. In this case we are slicing the molecule diagonally so we are left with [Rh(COD)Cl]. We need to break two Rh alkene bonds and one Rh-Cl bond. The Rh-Cl bond is the easiest. Chlorine is less electronegative than rhodium so it gets the electron pair. Cl- must mean Rh+ so Rh1+ contribution from this bond. Now with an alkene (or alkyne etc) you only have two options. Either the multiple bond is donating electron density to the metal centre or the other way around. We know that metals are electropositive, all more so than carbon according to the Pauling scale. Break this bond and you are left with COD and Rh. COD is a neutral species therefore it does not contribute to the formal oxidation state of rhodium. Rh+ + Rh0 = Rh1+ for this molecule. Here's a bunch of examples for you to scratch your head over to make sure you fully understand the process.